# Generating all divisors of a number using its prime factorization

Given an integer N, the task is to find all of its divisors using its prime factorization.

Examples:

Input: N = 6
Output: 1 2 3 6

Input: N = 10
Output: 1 2 5 10

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: As every number greater than 1 can be represented in it’s prime factorization as p1a1*p2a2*……*pkak, here pi is a prime number, k ≥ 1 and ai is a positive integer.
Now all the possible divisors can be generated recursively if the count of occurrence of every prime factor of n is known. For every prime factor pi, it can be included x times where 0 ≤ x ≤ ai. First, find the prime factorization of n using this approach and for every prime factor, store it with the count of it’s occurrence.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include "iostream" ` `#include "vector" ` `using` `namespace` `std; ` ` `  `struct` `primeFactorization { ` ` `  `    ``// to store the prime factor ` `    ``// and its highest power ` `    ``int` `countOfPf, primeFactor; ` `}; ` ` `  `// Recursive function to generate all the ` `// divisors from the prime factors ` `void` `generateDivisors(``int` `curIndex, ``int` `curDivisor, ` `                      ``vector& arr) ` `{ ` ` `  `    ``// Base case i.e. we do not have more ` `    ``// primeFactors to include ` `    ``if` `(curIndex == arr.size()) { ` `        ``cout << curDivisor << ``' '``; ` `        ``return``; ` `    ``} ` ` `  `    ``for` `(``int` `i = 0; i <= arr[curIndex].countOfPf; ++i) { ` `        ``generateDivisors(curIndex + 1, curDivisor, arr); ` `        ``curDivisor *= arr[curIndex].primeFactor; ` `    ``} ` `} ` ` `  `// Function to find the divisors of n ` `void` `findDivisors(``int` `n) ` `{ ` ` `  `    ``// To store the prime factors along ` `    ``// with their highest power ` `    ``vector arr; ` ` `  `    ``// Finding prime factorization of n ` `    ``for` `(``int` `i = 2; i * i <= n; ++i) { ` `        ``if` `(n % i == 0) { ` `            ``int` `count = 0; ` `            ``while` `(n % i == 0) { ` `                ``n /= i; ` `                ``count += 1; ` `            ``} ` ` `  `            ``// For every prime factor we are storing ` `            ``// count of it's occurenceand itself. ` `            ``arr.push_back({ count, i }); ` `        ``} ` `    ``} ` ` `  `    ``// If n is prime ` `    ``if` `(n > 1) { ` `        ``arr.push_back({ 1, n }); ` `    ``} ` ` `  `    ``int` `curIndex = 0, curDivisor = 1; ` ` `  `    ``// Generate all the divisors ` `    ``generateDivisors(curIndex, curDivisor, arr); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 6; ` ` `  `    ``findDivisors(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` `class` `GFG  ` `{ ` ` `  `static` `class` `primeFactorization  ` `{ ` ` `  `    ``// to store the prime factor ` `    ``// and its highest power ` `    ``int` `countOfPf, primeFactor; ` ` `  `    ``public` `primeFactorization(``int` `countOfPf,  ` `                              ``int` `primeFactor) ` `    ``{ ` `        ``this``.countOfPf = countOfPf; ` `        ``this``.primeFactor = primeFactor; ` `    ``} ` `} ` ` `  `// Recursive function to generate all the ` `// divisors from the prime factors ` `static` `void` `generateDivisors(``int` `curIndex, ``int` `curDivisor, ` `                           ``Vector arr) ` `{ ` ` `  `    ``// Base case i.e. we do not have more ` `    ``// primeFactors to include ` `    ``if` `(curIndex == arr.size())  ` `    ``{ ` `        ``System.out.print(curDivisor + ``" "``); ` `        ``return``; ` `    ``} ` ` `  `    ``for` `(``int` `i = ``0``; i <= arr.get(curIndex).countOfPf; ++i) ` `    ``{ ` `        ``generateDivisors(curIndex + ``1``, curDivisor, arr); ` `        ``curDivisor *= arr.get(curIndex).primeFactor; ` `    ``} ` `} ` ` `  `// Function to find the divisors of n ` `static` `void` `findDivisors(``int` `n) ` `{ ` ` `  `    ``// To store the prime factors along ` `    ``// with their highest power ` `    ``Vector arr = ``new` `Vector<>(); ` ` `  `    ``// Finding prime factorization of n ` `    ``for` `(``int` `i = ``2``; i * i <= n; ++i) ` `    ``{ ` `        ``if` `(n % i == ``0``) ` `        ``{ ` `            ``int` `count = ``0``; ` `            ``while` `(n % i == ``0``)  ` `            ``{ ` `                ``n /= i; ` `                ``count += ``1``; ` `            ``} ` ` `  `            ``// For every prime factor we are storing ` `            ``// count of it's occurenceand itself. ` `            ``arr.add(``new` `primeFactorization(count, i )); ` `        ``} ` `    ``} ` ` `  `    ``// If n is prime ` `    ``if` `(n > ``1``) ` `    ``{ ` `        ``arr.add(``new` `primeFactorization( ``1``, n )); ` `    ``} ` ` `  `    ``int` `curIndex = ``0``, curDivisor = ``1``; ` ` `  `    ``// Generate all the divisors ` `    ``generateDivisors(curIndex, curDivisor, arr); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args)  ` `{ ` `    ``int` `n = ``6``; ` ` `  `    ``findDivisors(n); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Recursive function to generate all the  ` `# divisors from the prime factors  ` `def` `generateDivisors(curIndex, curDivisor, arr): ` `     `  `    ``# Base case i.e. we do not have more  ` `    ``# primeFactors to include  ` `    ``if` `(curIndex ``=``=` `len``(arr)): ` `        ``print``(curDivisor, end ``=` `' '``) ` `        ``return` `     `  `    ``for` `i ``in` `range``(arr[curIndex][``0``] ``+` `1``): ` `        ``generateDivisors(curIndex ``+` `1``, curDivisor, arr)  ` `        ``curDivisor ``*``=` `arr[curIndex][``1``] ` `     `  `# Function to find the divisors of n  ` `def` `findDivisors(n): ` `     `  `    ``# To store the prime factors along  ` `    ``# with their highest power  ` `    ``arr ``=` `[] ` `     `  `    ``# Finding prime factorization of n  ` `    ``i ``=` `2` `    ``while``(i ``*` `i <``=` `n): ` `        ``if` `(n ``%` `i ``=``=` `0``): ` `            ``count ``=` `0` `            ``while` `(n ``%` `i ``=``=` `0``): ` `                ``n ``/``/``=` `i  ` `                ``count ``+``=` `1` `                 `  `            ``# For every prime factor we are storing  ` `            ``# count of it's occurenceand itself.  ` `            ``arr.append([count, i])  ` `     `  `    ``# If n is prime  ` `    ``if` `(n > ``1``): ` `        ``arr.append([``1``, n])  ` `     `  `    ``curIndex ``=` `0` `    ``curDivisor ``=` `1` `     `  `    ``# Generate all the divisors  ` `    ``generateDivisors(curIndex, curDivisor, arr)  ` ` `  `# Driver code  ` `n ``=` `6` `findDivisors(n)  ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

## C#

 `// C# implementation of the approach ` `using` `System;  ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `public` `class` `primeFactorization  ` `{ ` ` `  `    ``// to store the prime factor ` `    ``// and its highest power ` `    ``public` `int` `countOfPf, primeFactor; ` ` `  `    ``public` `primeFactorization(``int` `countOfPf,  ` `                              ``int` `primeFactor) ` `    ``{ ` `        ``this``.countOfPf = countOfPf; ` `        ``this``.primeFactor = primeFactor; ` `    ``} ` `} ` ` `  `// Recursive function to generate all the ` `// divisors from the prime factors ` `static` `void` `generateDivisors(``int` `curIndex, ``int` `curDivisor, ` `                             ``List arr) ` `{ ` ` `  `    ``// Base case i.e. we do not have more ` `    ``// primeFactors to include ` `    ``if` `(curIndex == arr.Count)  ` `    ``{ ` `        ``Console.Write(curDivisor + ``" "``); ` `        ``return``; ` `    ``} ` ` `  `    ``for` `(``int` `i = 0; i <= arr[curIndex].countOfPf; ++i) ` `    ``{ ` `        ``generateDivisors(curIndex + 1, curDivisor, arr); ` `        ``curDivisor *= arr[curIndex].primeFactor; ` `    ``} ` `} ` ` `  `// Function to find the divisors of n ` `static` `void` `findDivisors(``int` `n) ` `{ ` ` `  `    ``// To store the prime factors along ` `    ``// with their highest power ` `    ``List arr = ``new` `List(); ` ` `  `    ``// Finding prime factorization of n ` `    ``for` `(``int` `i = 2; i * i <= n; ++i) ` `    ``{ ` `        ``if` `(n % i == 0) ` `        ``{ ` `            ``int` `count = 0; ` `            ``while` `(n % i == 0)  ` `            ``{ ` `                ``n /= i; ` `                ``count += 1; ` `            ``} ` ` `  `            ``// For every prime factor we are storing ` `            ``// count of it's occurenceand itself. ` `            ``arr.Add(``new` `primeFactorization(count, i )); ` `        ``} ` `    ``} ` ` `  `    ``// If n is prime ` `    ``if` `(n > 1) ` `    ``{ ` `        ``arr.Add(``new` `primeFactorization( 1, n )); ` `    ``} ` ` `  `    ``int` `curIndex = 0, curDivisor = 1; ` ` `  `    ``// Generate all the divisors ` `    ``generateDivisors(curIndex, curDivisor, arr); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args)  ` `{ ` `    ``int` `n = 6; ` ` `  `    ``findDivisors(n); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```1 3 2 6
```

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